Answer:
<h2>
f₀ = 158.12 Hertz</h2>
Explanation:
The fundamental frequency of the string f₀ is expressed as f₀ = V/4L where V is the speed experienced by the string.
where T is the tension in the string and 
is the density of the string
Given T = 600N and = 0.015 g/cm = 0.0015kg/m
The next is to get the length L of the string. Since the string is stretched and fixed at both ends, 200 cm apart, then the length of the string in metres is 2m.
L = 2m
Substituting the derived values into the formula f₀ = V/2L
f₀ = 632.46/2(2)
f₀ = 632.46/4
f₀ = 158.12 Hertz
Hence the fundamental frequency of the string is 158.12 Hertz
F = mass × acceleration
17 = mass × 5
mass = 17/5
Answer:
(a) Force must be grater than 283.87 N
(B) Force will be equal to 193.945 N
Explanation:
We have given mass of the crate m = 49.6 kg
Acceleration due to gravity 
Coefficient of static friction 
Coefficient of kinetic friction 
(a) Static friction force is given by 
So to just start the crate moving we have to apply more force than 283.87 N
(B) This force will be equal to kinetic friction force
We know that kinetic friction force is given by 
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
